Evaluate the logarithm. Round your answer to the nearest thousandth. $2\log_{6}(33)\approx$
Solution: Thinking about the problem We know that the value of $\log_{6}(33)$ is between $ 1$ and $2$ since $6^1=6$ and $6^2=36$ and so ${2}\log_{6}(33)$ would be somewhere between ${2}\cdot 1=2$ and ${2}\cdot 2 = 4$. Given the numbers, though, it is difficult to approximate the answer accurate to three decimal places. In order to do this, we can use calculators. However, most calculators only calculate logarithms in base $10$ and base $e$. The change of base rule We can change the base of any logarithm by using the following rule: $\log_b(a)=\dfrac{\log_x(a)}{\log_x(b)}$, where $a>0$, $b>0$, $b\neq1$ and $x>0$, $x\neq1$. Changing to base- $10$ Let's change the base in this problem to $10$. $\begin{aligned}2\log_6(33)&=2\cdot\log_6(33)\\\\ &=2\cdot\left(\dfrac{\log_{10}(33)}{\log_{10}(6)}\right)\\\\ &=2\cdot\left(\dfrac{\log(33)}{\log(6)}\right)\\\\ &=\dfrac{2\log(33)}{\log(6)}\end{aligned}$ Evaluating with the calculator We can now approximate the value using the calculator. $\begin{aligned}\dfrac{2\log(33)}{\log(6)}\approx 3.903 \end{aligned}$ The answer is: $2\log_{6}(33)\approx 3.903$